# If an object is moving at 150 m/s over a surface with a kinetic friction coefficient of u_k=2250 /g, how far will the object continue to move?

May 28, 2016

$\Delta x = 5 m$

#### Explanation:

${v}_{i} = 150 \frac{m}{s} \text{ initial velocity}$
${u}_{k} = \frac{2250}{g} \text{ coefficient of friction}$

$\text{The object is decreased by the friction force}$

F_f=u_k*m*g"

${F}_{f} = \frac{2250}{\cancel{g}} \cdot m \cdot \cancel{g}$

$F = m \cdot a$

$2250 \cdot \cancel{m} = \cancel{m} \cdot a$

$a = 2250 \text{ } \frac{m}{s} ^ 2$

${v}_{f}^{2} = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$

$\text{the final velocity of object will be zero if object stops.}$

$0 = {v}_{i}^{2} - 2 \cdot a \cdot \Delta x$

$\Delta x = {v}_{i}^{2} / \left(2 \cdot a\right)$

$\Delta x = {150}^{2} / \left(2250\right)$

$\Delta x = \frac{22500}{2 \cdot 2250}$

$\Delta x = 5 m$